问题 解答题
(1)已知
1+
1
12
+
1
22
=
3
2
1+
1
22
+
1
32
=
7
6
1+
1
32
+
1
42
=
13
12
,…试猜测
1+
1
n2
+
1
(n+1)2
的结果,并加以证明;
1+
1
n2
+
1
(n+1)2
=
n2+n+1
n(n+1)

(2)s=
1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42
+…+
1+
1
20052
+
1
20062

求不超过S的最大整数[s].
答案

(1)猜想:

1+
1
n2
+
1
(n+1)2
=
n2+n+1
n(n+1)

证明:

1+
1
n2
+
1
(n+1)2
=
n2(n+1)2+(n+1)2+n2
n2(n+1)2
=
n2(n+1)2+2n(n+1)+1 
n(n+1)
=
n2+n+1
n(n+1)

(2)∵

n2+n+1
n(n+1)
=1+
1
n
-
1
n+1

∴s=1+1-

1
2
+1+
1
2
-
1
3
+1+
1
3
-
1
4
+…+1+
1
2005
-
1
2006
=2005+1-
1
2006
=2005
2005
2006

∴[s]=2005.

问答题
单项选择题